Diffusion of Gases. 473 



(y, yjr) that cross the section A per unit of time. It is 



Wj/(v) dv\v cos ty sin ty dijr 



+ n-ifiy) dv^ot cos 2 ty sin yjr dyjr. 



Of course if cos yjr is positive they enter the element of volume; 

 if cos yjr is negative they leave it. 



5. We next consider the momentum in direction x which is 

 carried across the section A by the class (v, yfr). If in be the 

 mass of a molecule, it is 



mn x f(v) dv{^v 2 COS 2 yfr + ^v* COS 3 yjr } sin \jr d^fr. 



If we integrate this expression according to \fr from it to 0, 

 we obtain the momentum in direction x which the whole class 

 v brings into our element of volume through the section A in 

 unit time, negative momentum carried out of the element by 

 molecules having cos yjr negative being treated as positive 

 momentum carried into the element. The result is 



mn if(y) dv -r I cos 2 yfr sin yjr d^fr, .... (A) 



the term in cos 3 yjr disappearing. 



6. We will now consider what happens at the other sec- 

 tion A'. 



The momentum in direction x of the class v which is carried 

 out of our element of volume at A' per unit of time is found 



from the expression A by writing n } + — - 8x, that is, 



dn ^ 



n l — - dx for %. It is therefore 



m ( n x — -T7 Sx j f(v) dv -^ I cos 2 ty sin -ty dty. . . ( A') 



%> o 



7. Comparing the expressions A and A', we see that the 

 momentum of the class v in the direction of positive x which 

 enters the element of volume in unit time exceeds the mo- 

 mentum which leaves the element of volume by the quantity 



m -j- &x -k/(v) dv I cos 2 yjr sin yjr dyjr; . 

 J o 



that i 



is, 



m ~B^ x ^ dv T 



If therefore there were no encounters, the class v within 

 our element of volume would gain x momentum per unit time 



